Lesson 32 Flashcards — Line of Sight and Wireless Channels

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1. Context only (not assessed): Name the four wireless propagation modes.

Surface wave, sky wave, forward scatter, and line of sight. Which you get is decided by frequency, not preference.

2. Context only (not assessed): Describe surface-wave propagation.

The wave hugs the Earth's surface by diffraction. Works at VLF, LF and MF (carrier \(f_c \lesssim 3\) MHz) and reaches beyond the horizon — used for AM broadcast and submarine communications.

3. Context only (not assessed): Describe sky-wave propagation.

The signal bounces off the ionosphere. Useful in HF (\(f_c \lesssim 30\) MHz); a larger angle of incidence — a more grazing ray — gives a better bounce and a longer hop. Used for long-haul radio and over-the-horizon radar, but it varies with time of day and solar activity.

4. Context only (not assessed): Describe forward scatter.

The signal strikes discontinuities in the troposphere and scatters in all directions; a small fraction reaches a distant receiver. Useful at VHF, UHF and SHF (from about 50 MHz up), very inefficient, but gives beyond-line-of-sight range without a satellite.

5. What is special about line-of-sight propagation?

All electromagnetic signals propagate by LOS, in every band. Its geometric limit is the curvature of the Earth; the other limit is power.

6. Give the LOS distance equation used in this course, with units.

\(r = \sqrt{2h}\) with \(h\) in feet and \(r\) in statute miles. The equation sheet writes it \(R_{LOS,mi}\) and \(h_{ft}\): the subscripts are the units.

7. Why is that equation 'improper', and why is it used anyway?

Because the units do not balance — the square root of feet should not give miles. It works because the 4/3 Earth radius model happens to make 5280 miles coincide with 5280 feet per mile. Engineers accept it because it is correct and fast.

8. Why is the Earth radius adjusted to 4/3 of its true value?

Because decreasing atmospheric pressure with altitude bends radio waves slightly downward, extending range beyond the geometric horizon.

9. How do you handle two elevated antennas?

\(R_{LOS,mi} = \sqrt{2h_{ft,1}} + \sqrt{2h_{ft,2}}\) — each antenna reaches its own horizon and the ranges add. An end on the surface has a height of zero and adds nothing.

10. Which conversions go with the LOS equation?

The equation sheet's two: 1 ft = 0.305 m and 1 mi = 1.61 km. Turn a height in meters into feet before the square root, and the answer in statute miles into kilometers after it.

11. A helicopter at 2000 ft searches for a ground radio. How close must it get?

\(R_{LOS,mi} = \sqrt{2(2000)} + 0 = \sqrt{4000} = 63.25\) miles, which is \(63.25 \times 1.61 = 101.8\) km.

12. Why does altitude have diminishing returns?

Range grows only as \(\sqrt{h}\) — quadrupling altitude only doubles range. You cannot build a tower out of the problem, which is why airborne relays and satellites exist.